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Question

At t=0, an arrow is fired vertically upwards with a speed of 100 msāˆ’1. A second arrow is fired vertically upwards with the same speed at t=5 s. Then (take g=10 msāˆ’2)(Neglect air friction)

A
the two arrows will be at the same height above the ground at t=12.50 s
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B
the two arrows will reach back their starting points at t=20 s and t=25 s respectively
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C
the ratio of the speeds of the first and second arrows at t=20 s will be 2:1
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D
the maximum height attained by either arrow will be 1000 m
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Solution

The correct option is C the ratio of the speeds of the first and second arrows at t=20 s will be 2:1
Let them meet at height h after time t.
h=100t12gt2 for first arrow
h=100(t5)12g(t5)2 for second arrow
t=12.5 (after solving). So (a) is correct.
Time of flight of first arrow: T=2ug=2×10010=20 s
Second arrow will reach after 5 s of reaching first, so (b) is correct
v1=10010×20=100 m/s
v2=10010×15=50 m/s
Ratio: v1v2=2:1, so (c) is correct.
Maximum height attained
H=u22g=(100)22×10=500 m.
Hence (d) is incorrect.

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